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Let $D$ denote a closed two dimensional figure as: $D=2+iT\to 2\to 2-\delta\to 2-\delta+i(T-\delta)\to \frac{1}{2}+\epsilon+i(T-\delta)\to\frac{1}{2}+\epsilon\to\frac{1}{2}-\epsilon\to \frac{1}{2}-\epsilon+iT\to 2+iT$ where $0<\epsilon<\frac{1}{2}$ and $T>3$. Assume that the Riemann zeta function, $\zeta(s)$ is non zero on the boundary of $D$ and there exists $\delta_0>0$ such that $\zeta(s)$ is non zero on and inside the $L-$ shaped region $L_\delta$, where $L_\delta=2+iT\to 2\to 2-\delta\to 2-\delta+i(T-\delta)\to \frac{1}{2}-\epsilon+i(T-\delta)\to\frac{1}{2}-\epsilon+iT\to 2+iT$ whenever $0<\delta<\delta_0$.

I need to apply Littlewood's lemma for $\zeta(s)$ in $D$.

Question: How do we define logarithm for all points $s$ in $D$ so that we can apply Littlewood's lemma for $\zeta(s)$ in $D$. Note that $D$ is not a rectangle so normal definition of logarithm as in the usual Littlewood's lemma might not work.

I tried the following: version of Littlewood's Lemma for $D$: Let $D$ be defined as above and $\zeta(s)$ be a holomorphic function in $D$ not vanishing on the boundary of $D$. Then \begin{equation}\sum_{\rho\in D} \text{dist}(\rho)=\frac{1}{2\pi i} \int_{\partial{D}}\log \zeta (s)\ ds \tag{1}\end{equation} where in the above integral we go through $\partial{D}$ in negative direction (clockwise), $\rho$ are the zeros of $\zeta(s)$ inside $D$ counted with multiplicity, $\text{dist}(\rho)$ denotes the distance of $\rho$ to the left side of $D$ and $\log \zeta(s)$ is defined by continuous variation as follows: We start with a particular determination on $\sigma=2$, and obtain the value at other points of $D$ by continuous variation from $\log \zeta(2+it)$ down to $\log \zeta(2)$ and then left to (a point in between $2$ and $2-\delta$) $\log \zeta(2a+(2-\delta)(1-a))$ (where $0\leq a\leq 1$) then upwards to $\log \zeta((2a+(2-\delta)(1-a))+it')$ (where $t'=(T-\delta)b+T(1-b)$ where $0\leq b\leq 1$) then left to $\log \zeta\left(\left(\frac{1}{2}-\epsilon\right)c+\left(\frac{1}{2}+\epsilon\right)(1-c)+it'\right)$ (where $0\leq c\leq 1$) and then down to $\log \zeta(\sigma+it)$ till we reach $\sigma+it$. Mathematically we have for $s\in D$ and $A=2a+(2-\delta)(1-a)$ , $B=\left(\frac{1}{2}-\epsilon\right)c+\left(\frac{1}{2}+\epsilon\right)(1-c)$ \begin{align}\log \zeta(\sigma+it)&=\log \zeta(2+it)\notag\\& +\left(\int_{2+it}^{2}+\int_{2}^{A}+\int_{A}^{A+it'}+\int_{A+it'}^{B+it'}+\int_{B+it'}^{\sigma+it}\right) \frac{\zeta'(s)}{\zeta(s)}\ ds \tag{2}\end{align} provided that $t$ and $t'$ are not the ordinate of zeros. If, however, this path would cross a zero or pole of $\zeta(s)$, we take $\log \zeta(s)$ to be $\log \zeta(s\pm i0)$ according as we approach the path from above or below.

Does the definition of logarithm needs any changes?

Thank you.

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  • $\begingroup$ Why the downvote? $\endgroup$
    – Honor
    Sep 8, 2022 at 16:01
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    $\begingroup$ I didn't downvote, but posting a wall of text with 12 (!) numbered equations and not much guide to what needs to be read to understand the question, and what is just background, definitely does encourage me to skip this question and move on to the next one. $\endgroup$
    – LSpice
    Sep 8, 2022 at 16:12
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    $\begingroup$ A proof of the Littlewood lemma, which answers your question, may be found here:web.williams.edu/Mathematics/sjmiller/public_html/ntandrmt/… $\endgroup$
    – Stopple
    Sep 8, 2022 at 16:45
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    $\begingroup$ The same proof works. You put little circles around all the zeros of $\zeta(s)$, with (two) line segments connecting each circle to the boundary of $D$ so $\log(\zeta(s))$ is well defined inside $D$ and outside the circles. Any choice of the branch is fine. $\endgroup$
    – Stopple
    Sep 8, 2022 at 17:07
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    $\begingroup$ as mentioned in comments/answers to many similar posts here and on MSE, the definition you want to use works only if a form of RH is assumed, namely that there are no zeroes of zeta in the strip $1/2+ \epsilon < \sigma < 1$ up to height $t$; the crucial thing in the proof of Littlewood lemma is that $\log F$ is analytic on the domain obtained when you remove the zeroes inside and segments joining them to the corresponding rectangle side you want to measure their distance from, while in the Titchmarsh definition we take out ALL the left half lines starting at ALL zeroes in the critical strip $\endgroup$
    – Conrad
    Sep 8, 2022 at 19:34

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